Question

If $\frac{si{n}^{4}x}{2}+\frac{co{s}^{4}x}{3}=\frac{1}{5}$ , prove that $ta{n}^{2}x=\frac{2}{3}$

Answer 1

We have, $\frac{si{n}^{4}x}{2}+\frac{co{s}^{4}x}{3}=\frac{1}{5}$

$\Rightarrow \frac{3si{n}^{4}x+2co{s}^{4}x}{6}=\frac{1}{5}$

$\Rightarrow 15si{n}^{4}x+10co{s}^{4}x=6$

On dividing both sides by $co{s}^{4}x$, we get

$\Rightarrow 15ta{n}^{4}x+10=6se{c}^{4}x$

$\Rightarrow 15ta{n}^{4}x+10=6(1+ta{n}^{2}x{)}^2[\because se{c}^2\theta -ta{n}^2\theta =1]$

$\Rightarrow 15ta{n}^{4}x+10=6+12ta{n}^{2}x+6ta{n}^{4}x$

$\Rightarrow 9ta{n}^{4}x-12ta{n}^{2}x+4=0$

$\Rightarrow (3ta{n}^{2}x-2{)}^2=0$

$\therefore ta{n}^{2}x=\frac{2}{3}\text{Hence proved.}$